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**1. The problTfem statement, all variables and given/known data**

Determine the result when 100g of steam at 100C is passed into a mixture of 200g of Water and 20g of ice at exactly 0C in a calorimeter which behaves thermally as if it were equvalent to 30g of water.

## Homework Equations

FInding the mass. I think I figured out the final temp?

## The Attempt at a Solution

Q=0=Steam condensed heat + Change in steam water heat + change in water heat + heat to melt Ice + change in calorimeter heat

(540cal/g)(100g) + (100g)(1cal/g C)(Tf-0C) +(200g)(1cal/g C)(Tf-0C) +(20g)(80g/C) +(20g)(1cal/g C)(Tf-0C) + (30g)(1cal/gC)(Tf-0C)

Tf=149.7

ITs all water so the temp has to be 100 it can't go higher than that.

I am struggling to find the grams of steam condensed.

m(-540g/C)=Ice melted + Rise in water?

Please help and thanks!